214 PHILOSOPHICAL TRANSACTIONS. [aNNO 1771- 



points of contact T, *, "¥, respectively. The area of the triangle, ahb, is equal 

 to the areas of the 3 sectors AT*, 8$^, H*T, added to the interstitial area 

 T*^. But the triangle ahb is equilateral. Therefore each of the sectors AT*, 

 B**', H^'T, is -^ of the circle to which it belongs : and the circles being equal, 

 the 3 sectors are equal to the half of any one of the circles. Therefore the area 

 of the triangle ahb is equal to 4- the area of one circle, (as of a) added to the 

 interstitial area T*Y. Therefore, from the area of the triangle ahb, take -J- the 

 area of the circle a, and there will remain the interstitial area T**. 



Now, if the radius A* be put = 1, each side of the triangle ahb will be 2. 



Therefore, the area of the triangle ahb will be = ] .73203 



But the radius being 1, -J- the area of the circle a is = l .5708 



The difference is 0.l6l2 



And this is the interstitial area T$*, the half area of the circle a being I.57OS. 

 Therefore the semicircle is to the interstice, as I.57O8 to O.1612, or as 9.74 to 

 1, or as 39 to 4, very nearly. 



Carol. If a parcel of equal circles be so disposed on a plane surface of any 

 figure whatever, that the centres of every 3 adjacent circles are situated at the 

 angles of equal equilateral triangles, having sides greater than the diameters of 

 the circles, but greater in a finite proportion, the ultimate proportion of the 

 space covered by all the circles, to the space occupied by all the interstices, when 

 each circle is infinitely diminished, and the number of them so infinitely in- 

 creased, that the space over which they spread is of a finite magnitude, is that 

 of -J- the area of one circle to the whole area of one interstice. And the area of 

 any one interstice, is equal to the difference of the area of the equilateral triangle, 

 formed by the right lines joining 3 adjacent centres, and 4- the area of one of 

 the circles. 



Frob. II. To determine the greatest possible density of an infinitely thin 

 crust composed of equal spherules, having their centres all in the same plane. 



From the number 39 subtract its third part. To the number 4 add the third 

 part of 39. The remainder is to the sum, that is, 26 is to 17, very nearly, as 

 the space occupied by ajHhe matter, to the space occupied by all the pore in an 

 infinitely thin crust, of the greatest possible density, composed of equal sphe- 

 rules, having all their centres in the same plane. 



Demonst. On a base of innumerable infinitely small circles, arranged in the 

 closest manner possible, according to prob. 1, imagine right cylinders to be 

 erected, each cylinder having one of the little circles for its base, and its altitude 

 equal to the diameter of its base. These cylinders are in the closest arrangement 

 possible for equal cylinders; and the spheres, which they circumscribe, are in 

 the closest arrangement possible for equal spheres, which have their centres in 

 the same plane. The solid space occupied by the cylinders, is to the solid space 



